The \(\mathrm{emf}, \mathrm{E}\), is related to the change in Gibbs free energy,
\(\Delta \mathrm{G}: \Delta \mathrm{G}=-\mathrm{nFE}\), where is the number of
electrons transferred during the redox process and \(F\) is a unit called the
Faraday. The faraday is the amount of charge on \(1 \mathrm{~mol}\) of
electrons: \(1 \mathrm{~F}=96,500 \mathrm{C} / \mathrm{mol}\). Because
\(\mathrm{E}\) is related to \(\Delta \mathrm{G}\), the sign of \(\mathrm{E}\)
indicates whether a redox process is spontaneous: \(\mathrm{E}>0\) indicates a
spontaneous process, and \(\mathrm{E}<0\) indicates a non-spontaneous one.
Given the cell:
\(\mathrm{Cd}(\mathrm{s})\left|\mathrm{Cd}(\mathrm{OH})_{2}(\mathrm{~s})\right|
\mathrm{NaOH}(\mathrm{aq}, 0.01 \mathrm{M})\)
\(\mid \mathrm{H}_{2}(\mathrm{~g}, 1\) bar \(\mid \mathrm{Pt}(\mathrm{s}))\)with
\(\mathrm{E}_{\text {cell }}=0.0 \mathrm{~V}\). If
\(\mathrm{E}_{\mathrm{Cd}}^{\circ}{ }_{\mathrm{Cd}}^{2+}=-0.39 \mathrm{~V}\),
then \(\mathrm{K}_{\mathrm{sp}}\) of
\(\mathrm{Cd}\left(\mathrm{OH}_{2}\right)\) is:
a. \(10^{-15}\)
b. \(10^{-13}\)
c. \(10^{13}\)
d. \(10^{15}\)